Set Theory/Intuitive Set Theory Let $A_1, A_2, ..., A_i$ be sets, and define $S_{x}$ to be $A_{1}\cup A_{2}\cup ... \cup A_{x}$ for $x=1, 2, 3, ..., r$.  Show that 
$\alpha=\{A_{1}, (A_{2}-S_{1}), ..., (A_{r}-S_{r-1})\}$
is a disjoint collection of sets and that
$S_{x}=A_{1}\cup(A_{2}-S_{1})\cup ... \cup (A_{x}-S_{x-1})$.
When is $\alpha$ a partition of $S_{x}$?

I have been trying to understand intuitive set theory and while doing some exercises, i came across this question in the section 'Algebra of Sets' and the author hadn't explained this section well.
I appreciate any hints given or clues.
Thank you.
 A: For the first part of the question, it may help to re-write it without the $S_{x}$ notation.
\begin{align*}
\alpha &= \{A_{1}, A_{2}-S_{1}, A_{3}-S_{2}, ..., A_{r}-S_{r-1}\} \\
       &= \{A_{1}, A_{2}-A_{1}, A_{3}-(A_{1}\cup A_{2}),..., A_{r}-(A_{1}\cup ... \cup A_{r-1})\}
\end{align*}
Clearly $A_{1}$ is disjoint from $A_{2}-A_{1}$.
If an element is in $A_{3}-(A_{1}\cup A_{2})$ then it is not in $A_{1}$ and it is not in $A_{2}$... so $A_{3}-(A_{1}\cup A_{2})$ is disjoint from $A_{1}$ and $A_{2}-A_{1}$, etc.  The thing to notice is that $S_{x-1}$ is the union of all the elements that could come in the sets appearing before $A_{x}$, so $A_{x}-S_{x-1}$ contains the elements that didn't come before.
So, after labor, is it "clear" that $\alpha$ is a collection of disjoint sets?  I think so.  You could of course proceed by induction if you really want to belabor the point... but that shouldn't really be necessary.
So, for the last question: When is $\alpha$ a partition of $S_{x}$, it may help to just think about what it means to be a partition.
